Question

Choose the correct option:
The electron configuration of Mg is:
A.\[{\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{d}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}\]
B.\[{\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}\]
C.\[{\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{d}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}\]
D.\[{\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^3}\]

Answer 1

Hint: Electronic configuration is just a representation of how the electrons are distributed in an atom. The electronic configuration of elements is governed by the Aufbau principle and Hund’s rule of maximum multiplicity. Mg or magnesium is an alkaline earth metal element. All the alkaline earth metals possess two electrons in their outermost shell.

Complete step by step answer:
According to the Aufbau principle, the electrons are filled in orbitals in order of their increasing energies. This means that an orbital with lower orbital energy will be filled first. Now, the energy content of any two subshells are compared by the ${\text{n + l}}$ rule, n is the principal quantum number and l is the orbital quantum number. According to the ${\text{n + l}}$ rule, the subshell with lower ${\text{n + l}}$ value will have lower energy and thus, will be filled first. If the ${\text{n + l}}$ values of both the subshells are the same, then the subshell with lower n value will have lower energy and will be filled first.
 For 1s orbital, n is equal to 1 and l is equal to 0, so, the ${\text{n + l}}$ value will be ${\text{1 + 0 = 1}}$ . Similarly, for 2s orbital, n is equal to 2 and l is equal to 0, so, the ${\text{n + l}}$ value will be ${\text{2 + 0 = 2}}$ , for 2p, n is equal to 2 and l is equal to 1, so, the ${\text{n + l}}$ value will be ${\text{2 + 1 = 3}}$ and for 3s, n is equal to 3 and l is equal to 0, so, the ${\text{n + l}}$ value will be ${\text{3 + 0 = 3}}$ . 2p and 3s have the same ${\text{n + l}}$ value, but 2p has lower n value than 3s. So, 2p has lower energy than 3s. Hence, the first 1s orbital will be filled, followed by 2s, 2p and then 3s.
According to Hund’s rule, an orbital cannot have more than 2 electrons and s and p subshells can have a maximum of 2 and 6 electrons respectively.
Magnesium is a group 2 element and has the atomic number 12. This means there are 12 electrons to be filled in magnesium. Now, following all the above mentioned rules, these 12 electrons will be filled. Proceeding in this way, we will have 2 electrons in 1s orbital followed by 2 electrons in 2s orbital, 6 electrons in 2p orbitals and 2 electrons in 3s orbital. Thus, this sums up 12 electrons. Hence, the electron configuration of Mg is \[{\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}\].

Hence option B is correct.

Note:
Some important characteristic of s-block elements are:
They are very soft and possess low melting and boiling points.
They also possess low ionization enthalpies.
They are highly reactive and easily form univalent or bivalent positive ions by losing the valence electrons.
The s- block metals and their corresponding salts impart characteristic colours to the flame.